Question

Given 13.5 grams of substance Z, if the substance absorbs 36 kilojoules of energy and the temperature increases by 25 degrees Celsius, what is the specific heat of the substance? 1.07 x 102 J/g·°C 1.22 x 104 J/g·°C 1.07 x 10-1 J/g·°C 1.22 x 107 J/g·°C

Answer 1

1.07 ×10² j/g.°C

Step-by-step explanation:

Given data:

Mass of substance Z = 13.5 g

Amount of heat absorbed = 36 Kj (36×1000 = 36000 j)

Increase in temperature= ΔT = 25°C

Specific heat of substance = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = m.c. ΔT

36000 j = 13.5 g ×c×25°C

36000 j / 13.5 g×25°C = c

c = 36000 j /337.5 g.°C

c = 1.07 ×10² j/g.°C